# Is this schema equivalent to Replacement under removal of Extensionality?

If $\phi(x)$ is a formula in which only symbol $x"$ occurs free, and it only occurs free, and in which symbol $y"$ never occurs; and if $\phi(y)$ is the formula obtained from $\phi(x)$ by merely replacing each occurrence of symbol $x"$ by symbol $y"$, then:

$\exists A \forall x,y (\phi(x) \wedge \phi(y) \to \exists m \in A (m \in x) \wedge [x \neq y \to \not \exists z (z \in x \wedge z \in y)]) \to \exists x \forall y (y \in x \leftrightarrow \phi(y))$

is an axiom.

In English any definable unary predicate $\phi$ for which there exists a set A such that $\phi$ predicates pairwise disjoint sets containing an element A among their elements, then $\phi$ defines a set!

Now I think that the above when added to axioms of Zermelo set theory $\text{Z}$ would prove all instances of the Replacement schema. In other words the above axiom schema can be regarded as a form of Replacement scheme, Call it $\text{Rpl}^*$

Question1: Can $\text{Z - Extensionality + Rpl}^*$ interpret full $\text{ZFC}$?

Question2: if we add the condition $\not \exists z \in A$ instead of just $\not \exists z$ in $\text{Rpl}^*$, then would that interpret full $\text{ZFC}$?

The idea is that if we have a schema that when added to $\text{Z-Extensionality}$ can prove replacement of elements of sets by co-extensional sets then this schema when added to Z-Extensionality would interpret full ZFC, a result due to Dana Scott. Formally this is:

$\forall x \in A \exists z \forall y (\phi(x,y) \to y \approx z) \to \exists B\forall y (y \in B \leftrightarrow \exists x \in A (\phi(x,y)))$

where $\phi(x,y)$ is a formula in which symbols $x,y"$ occur free, and $y \approx z$ stand for "$y$ is co-extensional to $z$" defined as: $\forall m (m \in y \leftrightarrow m \in z)$

any schema that can prove that scheme when added to $\text{Z-Extensionality}$ then it would interpret full $\text{ZFC}$. Dana Scott

Now the idea is that if we change $\neq$ symbol in $\text{Rpl}^*$ to $\not \approx$, and change $\not \exists z"$ by $\not \exists z \in A"$, then adding that to Z-Extensionality would (I think) prove the above schema of Dana Scott and so it would interpret Full ZFC.

But for the case of the question the problem I'm facing is that in absence of Extensionality there is no control on the amount of co-extensional copying of a set and so this might hamper proving Dana Scott's schema mentioned above, so the question is largely if there is a way to circumvent that?

• Can you add a reference for the Dana Scott result? – Joel David Hamkins Jul 2 '18 at 11:08
• Thanks. I see that Scott is relying on replacement, rather than collection, which perhaps explains the result. When you lack extensionality, then it is as though you can have many copies of the same set, which will tend to make many instances of replacement vacuous, but not collection. – Joel David Hamkins Jul 2 '18 at 12:09
• I'll try to present a proof of my alleged claim here that I wrote in the lines immediately following the reference of Dana Scott. Perhaps a modification of it would work to answer the second question, but the first question I believe is of the same strength of ordinary replacement. – Zuhair Al-Johar Jul 2 '18 at 13:20